证:a²+b²+c²=ab+bc+ca2a²+2b²+2c²-2ab-2bc-2ca=0a²-2ab+b²+b²-2bc+c²+c²-2ca+a²=0(a-b)²+(b-c)²+(c-a)²=0平方项恒非负,和=0,则各项均=0a-b=0 a=bb-c=0 b=cc-a=0 c=aa=b=c